Question:
An operator manufactures 10 identical spur gears in a lot. One spur gear is defective. Three spur gears are drawn at random without replacement. The probability of getting all three gears as non-defective is \(\underline{\hspace{2cm}}\). [round off to two decimal places]
An operator manufactures 10 identical spur gears in a lot. One spur gear is defective. Three spur gears are drawn at random without replacement. The probability of getting all three gears as non-defective is \(\underline{\hspace{2cm}}\). [round off to two decimal places]
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For "all good items" in sampling without replacement, use combinations: $\binom{\text{good}}{\text{drawn}}/\binom{\text{total}}{\text{drawn}}$.
Updated On: Jan 13, 2026
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Correct Answer: 0.69 - 0.71
Solution and Explanation
There are 10 gears:
- Non-defective = 9
- Defective = 1
We draw 3 gears without replacement. Probability that all 3 selected gears are good: \[ P = \frac{\binom{9}{3}}{\binom{10}{3}} \] \[ \binom{9}{3} = 84, \binom{10}{3} = 120 \] \[ P = \frac{84}{120} = 0.70 \] Thus the probability lies in the range \[ \boxed{0.69\text{ to }0.71} \]
- Non-defective = 9
- Defective = 1
We draw 3 gears without replacement. Probability that all 3 selected gears are good: \[ P = \frac{\binom{9}{3}}{\binom{10}{3}} \] \[ \binom{9}{3} = 84, \binom{10}{3} = 120 \] \[ P = \frac{84}{120} = 0.70 \] Thus the probability lies in the range \[ \boxed{0.69\text{ to }0.71} \]
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